If #2^x xx4^(x+1)=8# what is the value of #x#?

Question

Madelyn Anderson · Accepted Answer

#x = 1/3# #2^x * 4^(x+1)=8# #2^x * (2^2)^(x+1)=2^3# #2^x * 2^(2(x+1))=2^3# #2^(x + 2x+2)=2^3# #2^(color(red)((3x+2 )))=2^(color(red)(3))# #3x + 2 = 3# #x = 1/3#

Lily Anderson · Answer

High detail using first principles. Plus an alternative approach for the end.

#x=1/3# Given:#" "2^x xx4^(x+1)=8# But 4 is #2^2# so we have: #2^x xx(2^2)^(x+1)=8# #2^x xx2^(2(x+1))=8# #2^x xx2^(2x+2)=8# But #2^(2x+2)# is the same as: #2^(2x) xx2^2# giving #2^x xx2^(2x)xx2^2=8# #2^(3x) xx4=8# Split each side in half. #2^(3x) xx4xx1/4=8xx1/4# #2^(3x)=2^1" " rarr" "3x=1=>x=1/3# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Alternative approach")# You can solve the above problem directly, but I'd like to show you another way that might work in a different situation: take loges of both sides remembering that #log(2^(3x))->3xlog(2)# #3xlog(2)=log(2)# Divide both sides by #log(2)# #3x=1# divide each side by three. #x=1/3#

Sadie Alexander · Answer

undefined To find the value of $ x $, we can start by simplifying the equation $ 2^x \times 4^{x+1} = 8 $. Since $ 4 $ can be expressed as $ 2^2 $, we can rewrite $ 4^{x+1} $ as $ (2^2)^{x+1} $. Using the property of exponents, we get $ 4^{x+1} = 2^{2(x+1)} $. Substituting this back into the original equation, we have $ 2^x \times 2^{2(x+1)} = 8 $. Now, using the property of exponents for multiplication, we add the exponents when the bases are the same, resulting in $ 2^{x+2x+2} = 8 $. Simplifying the exponent, we have $ 2^{3x+2} = 8 $. Since $ 8 = 2^3 $, we can rewrite the equation as $ 2^{3x+2} = 2^3 $. Setting the exponents equal to each other, we get $ 3x+2 = 3 $. Solving for $ x $, we subtract $ 2 $ from both sides to get $ 3x = 1 $, then divide both sides by $ 3 $ to find $ x = \frac{1}{3} $. So, the value of $ x $ is $ \frac{1}{3} $.